What is the ball's height at the lowest point of its trajectory?
Consider a bungee cord of unstretched length
L0 = 48 m. When the cord is stretched to
L > L0 it behaves like a spring and obeys
Hooke’s law with the spring constant k =38 N/m. However, unlike a spring, the cord
folds instead of becoming compressed when
the distance between its ends is less than the
unstretched length: For L < L0 the cord has
zero tension and zero elastic energy.
To test the cord’s reliability, one end is tied
to a high bridge (height H = 138 m above the
surface of a river) and the other end is tied to
a steel ball of weight mg = 84 kg × 9.8 m/s
The ball is dropped off the bridge with zero
initial speed. Fortunately, the cord works and
the ball stops in the air before it hits the water
— and then the cord pulls it back up.
Calculate the ball’s height hbot at the lowest
point of its trajectory. For simplicity, neglects
the cord’s own weight and inertia as well as
the air drag on the ball and the cord.
Answer in units of m
What would the upward acceleration of the ball at the lowest point be?
- NCSLv 78 months agoFavourite answer
The ball's GPE is converted into EPE in the "spring":
m*g*(H - hbot) = ½kx²
where x = spring extension = H - hbot - L0
84kg*9.8m/s²*(138m - hbot) = ½*38N/m*(138m - hbot - 48m)²
This quadratic has solutions at
hbot = 119 m ← which is actually "htop"
and hbot = 17.8 m ≈ 18 m ◄ solution
Hope this helps!